Question

-what is the magnitude of tension in the rope? -find the horizontal and vertical components of the force exerted at A on the lamina by the wall.

A uniform triangular lamina ABC is right-angled at B and has sides
AB = 0.6 m and BC = 0.8 m. The mass of the lamina is 4 kg. One end of a light
inextensible rope is attached to the lamina at C. The other end of the rope is
attached to a fixed point D on a vertical wall. The lamina is in equilibrium
with A in contact with the wall at a point vertically below D. The lamina is in
a vertical plane perpendicular to the wall, and AB is horizontal. The rope is
taut and at right angles to AC

Find
(i) the tension in the rope,
(ii) the horizontal and vertical components of the force exerted at A on the
lamina by the wall.

Answer 1

See below.

Explanation:

For a homogeneous lamina, the triangle's center of mass is given by

`G = 1/3(A+B+C)`

Now momentum equilibrium

`vec M_A = (C-A) xx vec V +(G-A) xx vec P = vec 0`

and static resultant equilibrium

`vec R=vec V + vec H + vec P = vec 0`

Here

`A = (0,0)`
`B = (0.6,0)`
`C = (0.6,0.8)`
`G = (0.4,0.26667)` Lamina center of mass.
`vec V = v(-costheta_0, sintheta_0)` Rope tension force.
`vec H = (h_x,h_y)` Reaction force at point `A`.
`vec P = mg(0,-1)` Lamina weight .

`theta_0 = arctan(bar(CB)/bar(AB))`

After substituting values on `vec M_A = vec 0` and `vec R = vec 0` we obtain

`{(0.96 v - 16 = 0),(h_x-0.6v=0),(h_y+0.8v=40):}`

giving

`v = 16.6667`
`h_x = 10`
`h_y = 26.6667`

or

`vec V = 11.6667(-costheta_0, sintheta_0)`
`vec H = (10,26.6667)`